One possible argument is to compare this with $T/C(Z(U))$ as follows. We have $$|C(E):C(E)\cap C(Z(U))|=|C(E)C(Z(U)):C(Z(U))|$$ $$\leqslant|T:C(Z(U))|\leqslant|T:U\Phi(T)|$$ (since $\Phi(T)$ is central in $T$, it centralises $Z(U)$). Now $T/U$ is cyclic and $T/\Phi(T)$ has exponent $p$, so $T/U\Phi(T)$ is cyclic of exponent dividing $p$, so has order $1$ or $p$.

One possible argument is to compare this with $T/C(Z(U))$ as follows. We have $$C(E)/C(E)\cap C(Z(U))\cong C(E)C(Z(U))/C(Z(U)) \leqslant T/C(Z(U))$$ Now $T/U$ is cyclic, so $T/C(Z(U))$ is cyclic. It acts faithfully on $Z(U)$ by conjugation, and if $x$ is an element of $T$ whose image is a generator of $T/C(Z(U))$ then $[x,Z(U)]\leqslant T'$ (since $T/T'$ is abelian), so $x^p$ centralises $Z(U)$ since $T'$ has order $p$. Thus $x^p\in C(Z(U))$, which implies that $T/C(Z(U))$ has order $1$ or $p$.